= Solution
Because $w_k\geq c>0$, a finite sum $\sum_kw_k|u_k|$ implies $u\in\ell^1$. Thus the printed <weighted one-norm on l2> equals the extended nonnegative sum on every $u\in\ell^2$. Unbounded weights may make that sum infinite for some $u\in\ell^1$; membership in $\ell^1$ alone is not a finiteness assertion.
For each $N$, $J_N(u)=\sum_{k=1}^Nw_k|u_k|$ is continuous in the <l2 sequence space>, and $J=\sup_NJ_N$. If $u^{(n)}\to u$ in <norm>, then for every $N$,
$$
J_N(u)=\lim_nJ_N(u^{(n)})\leq\liminf_nJ(u^{(n)}).
$$
Taking the supremum over $N$ proves
$$
\boxed{J(u)\leq\liminf_nJ(u^{(n)}).}
$$
It is also weakly lower semicontinuous by the same finite-coordinate argument, and convex and proper because zero and finite sequences have finite penalty.
At a finite-penalty base point $v$, coordinate variations give the <subdifferential of a weighted one-norm on l2>:
$$
p\in\partial J(v)\iff p\in\ell^2,\quad p_k=w_k\operatorname{sign}(v_k)\ (v_k\ne0),\quad |p_k|\leq w_k\ (v_k=0).
$$
These conditions are sufficient by summing the coordinate supporting inequalities. Since $w_k\geq c$, an infinitely supported $v$ would require infinitely many $|p_k|\geq c$, contradicting $p\in\ell^2$. Thus a finite-penalty point can have empty <subdifferential>; a Bregman base with an actual <subgradient> must be finitely supported.
For any such $p$, $\langle p,v\rangle=J(v)$ and the <Bregman distance> simplifies to
$$
\boxed{D_J^p(u,v)=J(u)-\langle p,u\rangle=\sum_{k\geq1}(w_k|u_k|-p_ku_k).}
$$
Each summand is nonnegative; the sum may be infinite. The pairing converges by <Cauchy-Schwarz inequality>. This definition does not silently assume a <subgradient> exists at every $v\in\ell^1$.
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